Solutions Manual
for
Heat and Mass Transfer: Fundamentals & Applications
Fourth Edition
Yunus A. Cengel & Afshin J. Ghajar
McGraw-Hill, 2011
Chapter 13
RADIATION HEAT TRANSFER
PROPRIETARY AND CONFIDENTIAL
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View Factors
13-1C The view factor represents the fraction of the radiation leaving surface i that strikes surface j directly. The view factor from a surface to itself is non-zero for concave surfaces.
j i
F→
13-2C The cross-string method is applicable to geometries which are very long in one direction relative to the other directions. By attaching strings between corners the Crossed-Strings Method is expressed as
i Fi j surface on string 2 strings Uncrossed strings Crossed × − =
∑
∑
→13-3C The summation rule for an enclosure and is expressed as where N is the number of surfaces of the enclosure. It states that the sum of the view factors from surface i of an enclosure to all surfaces of the enclosure, including to itself must be equal to unity.
∑
= → = N j j i F 1 1The superposition rule is stated as the view factor from a surface i to a surface j is equal to the sum of the view factors from surface i to the parts of surface j, F1→(2,3) =F1→2+F1→3.
13-4C The pair of view factors and are related to each other by the reciprocity rule where Ai is the area of the surface i and Aj is the area of the surface j. Therefore,
j i F→ Fj→i AiFij =AjFji 21 1 2 12 21 2 12 1 F A A F F A F A = ⎯⎯→ =
13-5 An enclosure consisting of five surfaces is considered. The number of view factors this geometry involves and the number of these view factors that can be determined by the application of the reciprocity and summation rules
are to be determined. 2
3 5
4 1
Analysis A five surface enclosure (N=5) involves view
factors and we need to determine
25 = = 2 2 5 N 10 2 ) 1 5 ( 5 2 ) 1 (N− = − = N view factors directly. The remaining 25-10 = 15 of the view factors can be determined by the application of the reciprocity and summation rules.
13-6 An enclosure consisting of thirteen surfaces is considered. The number of view factors this geometry involves and the number of these view factors that can be determined by the application of the reciprocity and summation rules are to be
determined. 2 1 3 9 11 13 10 4 5 8 6 7 12 Analysis A thirteen surface enclosure (N = 13) involves
view factors and we need to determine
169 = = 2 2 13 N 78 2 ) 1 13 ( 13 2 ) 1 ( − = = − N
N view factors directly. The
remaining 169 - 78 = 91 of the view factors can be determined by the application of the reciprocity and summation rules.
13-7 A cylindrical enclosure is considered. (a) The expression for the view factor between the base and the side surface F13
in terms of K and (b) the value of the view factor F13 for L = D are to be determined.
Assumptions 1 The surfaces are diffuse emitters and reflectors.
Analysis (a) The surfaces are designated as follows: Base surface as A1, top surface as A2, and side surface as A3
Applying the summation rule to A1, we have
(where 1 13 12 11+F +F = F F11=0) or F13 =1 F− 12 (1)
For coaxial parallel disks, from Table 13-1, with i = 1, j = 2, ] ) 4 ( [ 2 1 4 2 1 2 1/2 2 / 1 2 1 2 2 12 = − − ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − = S S D D S S F (2) 2 2 2 2 1 2 2 2 4 ) / ( 4 2 1 2 1 1 K L D R R R S = + + = + = + = + (3) where K L D R R R 2 1 2 2 1= = = =
Substituting Eq. (3) into Eq. (2), we get
2 / 1 2 2 2 / 1 2 2 2 / 1 2 4 2 2 / 1 2 2 2 12 ) 1 ( 2 2 1 ] ) 1 ( 4 4 2 [ 2 1 ] ) 16 16 ( 4 2 [ 2 1 } ] 4 ) 4 2 [( 4 2 { 2 1 + − + = + − + = + − + = − + − + = K K K K K K K K K K K F
Substituting the above expression for F12 into Eq. (1) yields the expression for F13:
] ) 1 ( 2 2 1 [ 1 2 2 1/2 13= − + K − K K + F Hence, 2 2 / 1 2 13 2K(K 1) 2K F = + −
(b) The value of the view factor F13 for L = D (i.e., K = 1) is 0.828 = − = − + =2(1)(12 1)1/2 2(1)2 2 2 2 13 F
Discussion If the cylinder has a length and diameter of L = 2D, then from the expression for F13 we have
0.944 = − + = 2 1/2 2 13 2(2)(2 1) 2(2) F
13-8 The expression for the view factor F12 of two infinitely long parallel plates is to be determined using the Hottel’s
crossed-strings method.
Assumptions 1 The surfaces are diffuse emitters and reflectors. Analysis From the Hottel’s crossed-strings method, we have
) surface on String ( 2 ) strings Uncrossed ( ) strings Crossed ( i Fi j × Σ − Σ = →
For uncrossed strings, we have
w w w w w L L1= 2 =( 2+ 2)1/2 =( 2 + 2)1/2 = 2
For crossed strings, we have
w w
w
L ( 2 4 2)1/2 5
3 = + = and L4 =w
Applying the Hottel’s crossed-strings method, we get F12 as
0.204 = + − + = + − + = w w w w w w L L L L F 2 ) 2 2 ( ) 5 ( 2 ) ( ) ( 3 4 1 2 12
Discussion The Hottel’s crossed-string method is applicable only to surfaces that are very long, such that they can be considered to be two-dimensional and radiation interaction through the end surfaces is negligible.
13-9 The view factors between the rectangular surfaces shown in the figure are to be determined. Assumptions The surfaces are diffuse emitters and reflectors.
Analysis From Fig. 13-6,
W = 4 m 26 . 0 3 . 0 4 2 . 1 3 . 0 4 2 . 1 31 1 3 = ⎪ ⎪ ⎭ ⎪⎪ ⎬ ⎫ = = = = F W L W L (1) A2 A1 A3 (3) L3 = 1.2 m L2 = 1.2 m L1 = 1.2 m (2) and 32 . 0 6 . 0 4 4 . 2 3 . 0 4 2 . 1 ) 2 1 ( 3 2 1 3 = ⎪ ⎪ ⎭ ⎪⎪ ⎬ ⎫ = = + = = + → F W L L W L
We note that A1 = A3. Then the reciprocity and superposition rules gives 0.26 = = ⎯→ ⎯ = 3 31 13 31 13 1 A F A F F F 06 . 0 26 . 0 32 . 0 32 32 32 31 ) 2 1 ( 3→ + =F +F ⎯⎯→ = +F ⎯⎯→F = F Finally, A2=A3 ⎯⎯→F23 =F32=0.06
13-10 A cylindrical enclosure is considered. The view factor from the side surface of this cylindrical enclosure to its base surface is to be determined.
Assumptions The surfaces are diffuse emitters and reflectors. Analysis We designate the surfaces as follows:
Base surface by (1),
D=2r
(1) (3)
(2) top surface by (2), and
side surface by (3). Then from Fig. 13-7
L=2D=4r 05 . 0 25 . 0 4 4 4 21 12 2 2 2 1 1 1 = = ⎪ ⎪ ⎭ ⎪⎪ ⎬ ⎫ = = = = F F r r L r r r r L 1 : rule summation F11+F12+F13 = 95 . 0 1 05 . 0 0+ +F13 = ⎯⎯→F13 = 0.119 = = = = = ⎯→ ⎯ = (0.95) 8 1 8 2 : rule y reciprocit 2 13 1 2 1 13 1 2 1 13 3 1 31 31 3 13 1 F r r F L r r F A A F F A F A π π π π Discussion This problem can be solved more accurately by using the view factor relation from Table 13-1 to be
25 . 0 4 25 . 0 4 2 2 2 2 1 1 1 1 = = = = = = r r L r R r r L r R 056 . 0 1 1 4 18 18 4 18 25 . 0 25 . 0 1 1 1 1 5 . 0 2 2 2 1 5 . 0 2 1 2 2 2 1 12 2 2 2 1 2 2 = ⎪⎭ ⎪ ⎬ ⎫ ⎪⎩ ⎪ ⎨ ⎧ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − − = ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − = = + + = + + = R R S S F R R S 944 . 0 056 . 0 1 1 12 13 = −F = − = F 0.118 = = = = = ⎯→ ⎯ = (0.944) 8 1 8 2 : rule y reciprocit 2 13 1 2 1 13 1 2 1 13 3 1 31 31 3 13 1 F r r F L r r F A A F F A F A π π π π
13-11 A semispherical furnace is considered. The view factor from the dome of this furnace to its flat base is to be determined.
Assumptions The surfaces are diffuse emitters and reflectors.
(1) (2)
D
Analysis We number the surfaces as follows: (1): circular base surface
(2): dome surface
Surface (1) is flat, and thus F11=0.
1 1 : rule Summation F11+F12 = →F12 = 0.5 = = = = = ⎯→ ⎯ = 2 1 2 4 ) 1 ( A : rule y reciprocit 2 2 2 1 12 2 1 21 21 2 12 1 D D A A F A A F F A F π π
13-12 Three infinitely long cylinders are located parallel to each other. The view factor between the cylinder in the middle and the surroundings is to be
determined.
Assumptions The cylinder surfaces are diffuse emitters and reflectors.
(surr) D D s (1) (2) s D (2) Analysis The view factor between two cylinder facing each other
is, from Prob. 13-17,
D s D s F π ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − = − 2 2 2 1 2
Noting that the radiation leaving cylinder 1 that does not strike the cylinder will strike the surroundings, and this is also the case for the other half of the cylinder, the view factor between the cylinder in the middle and the surroundings becomes
D s D s F F surr π ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − − = − = − − 2 2 2 1 1 4 1 2 1
13-13 The view factors from the base of a cube to each of the other five surfaces are to be determined.
(2) (3), (4), (5), (6)
side surfaces Assumptions The surfaces are diffuse emitters and reflectors.
Analysis Noting that L1/D=L2/D=1, from Fig. 13-6 we read 2 . 0 12 = F
Because of symmetry, we have
0.2 = = = = = 13 14 15 16 12 F F F F F (1)
13-14 The view factor from the conical side surface to a hole located at the center of the base of a conical enclosure is to be determined.
Assumptions The conical side surface is diffuse emitter and reflector.
(3) h
(2) (1) d D
Analysis We number different surfaces as
the hole located at the center of the base (1) the base of conical enclosure (2) conical side surface (3)
Surfaces 1 and 2 are flat, and they have no direct view of each other. Therefore,
0 21 12 22 11 =F =F =F = F 1 1 : rule summation F11+F12 +F13 = ⎯⎯→F13 = 2Dh d2 = ⎯→ ⎯ = ⎯→ ⎯ = 3 31 2 31 31 13 1 4 (1) 2 A : rule y reciprocit F A F πd πDhF F
13-15 The four view factors associated with an enclosure formed by two very long concentric cylinders are to be determined. Assumptions 1 The surfaces are diffuse emitters and reflectors. 2 End effects are neglected.
Analysis We number different surfaces as
(2) (1)
D2 D1
the outer surface of the inner cylinder (1) the inner surface of the outer cylinder (2) No radiation leaving surface 1 strikes itself and thusF11=0
All radiation leaving surface 1 strikes surface 2 and thus F12 =1
2 1 D D = = = ⎯→ ⎯ = (1) A : rule y reciprocit 2 1 12 2 1 21 21 2 12 1 h D h D F A A F F A F π π 2 1 D D 1− = − = ⎯→ ⎯ = + 22 22 21 21 1 1 : rule summation F F F F
13-16 The view factors between the rectangular surfaces shown in the figure are to be determined. Assumptions The surfaces are diffuse emitters and reflectors.
Analysis We designate the different surfaces as follows: 4 m
shaded part of perpendicular surface by (1),
(4) 1 m (2) 1 m 1 m 1 m (1) (3) bottom part of perpendicular surface by (3),
shaded part of horizontal surface by (2), and front part of horizontal surface by (4).
(a) From Fig.13-6
26 . 0 25 . 0 4 1 25 . 0 4 1 23 1 2 = ⎪ ⎪ ⎭ ⎪⎪ ⎬ ⎫ = = = = F W L W L and 0.33 25 . 0 4 1 5 . 0 4 2 ) 3 1 ( 2 1 2 = ⎪ ⎪ ⎭ ⎪⎪ ⎬ ⎫ = = = = + → F W L W L 07 . 0 26 . 0 33 . 0 : rule ion superposit F2→(1+3) =F21+F23 ⎯⎯→F21=F2→(1+3) −F23 = − = 0.07 = = ⎯→ ⎯ = ⎯→ ⎯ = 2 1 12 2 21 12 21 1 : rule y reciprocit A A AF A F F F (b) From Fig.13-6, 0.16 5 . 0 4 2 0.25 4 1 3 ) 2 4 ( 1 2 = ⎪ ⎪ ⎭ ⎪⎪ ⎬ ⎫ = = = = → + F W L W L and 0.24 5 . 0 4 2 0.5 4 2 ) 3 1 ( ) 2 4 ( 1 2 = ⎪ ⎪ ⎭ ⎪⎪ ⎬ ⎫ = = = = + → + F W L W L 08 . 0 16 . 0 24 . 0 : rule ion superposit F(4+2)→(1+3) =F(4+2)→1+F(4+2)→3 ⎯⎯→F(4+2)→1= − = 16 . 0 ) 08 . 0 ( 4 8 : rule y reciprocit 1 ) 2 4 ( 1 ) 2 4 ( ) 2 4 ( 1 ) 2 4 ( 1 1 1 ) 2 4 ( ) 2 4 ( = = = ⎯→ ⎯ = → + + + → + → → + + F A A F F A F A 4 m (4) (2) 1 m 1 m 1 m 1 m (1) (3) 0.08 = − = ⎯→ ⎯ + = + → 08 . 0 16 . 0 : rule ion superposit 14 12 14 ) 2 4 ( 1 F F F F
since = 0.07 (from part a). Note that in part (b) is equivalent to in part (a).
F12 F14
12
(c) We designate
shaded part of top surface by (1), 3 m
3 m 1 m (3) 1 m 1 m 1 m (4) (2) (1) remaining part of top surface by (3),
remaining part of bottom surface by (4), and shaded part of bottom surface by (2).
From Fig.13-5, 15 . 0 67 . 0 3 2 1 3 3 ) 3 1 ( ) 4 2 ( 1 2 = ⎪ ⎪ ⎭ ⎪⎪ ⎬ ⎫ = = = = + → + F D L D L and 0.082 33 . 0 3 1 1 3 3 14 1 2 = ⎪ ⎪ ⎭ ⎪⎪ ⎬ ⎫ = = = = F D L D L 3 ) 4 2 ( 1 ) 4 2 ( ) 3 1 ( ) 4 2 ( : rule ion superposit F + → + =F + → +F + → 3 ) 4 2 ( 1 ) 4 2 ( : rule symmetry F + → = F + →
Substituting symmetry rule gives
075 . 0 2 15 . 0 2 ) 3 1 ( ) 4 2 ( 3 ) 4 2 ( 1 ) 4 2 ( + → = + → = + → + = = F F F 15 . 0 ) 075 . 0 )( 4 ( ) 2 ( : rule y reciprocit A1F1→(2+4) =A(2+4)F(2+4)→1⎯⎯→ F1→(2+4)= ⎯⎯→F1→(2+4)= 0.068 = − = ⎯→ ⎯ + = ⎯→ ⎯ + = + → 0.15 0.082 0.15 0.082 : rule ion superposit F1 (2 4) F12 F14 F12 F12
13-17 The view factor between the two infinitely long parallel cylinders located a distance s apart from each other is to be determined.
Assumptions The surfaces are diffuse emitters and reflectors.
D
(1) (2)
s
D
Analysis Using the crossed-strings method, the view factor between two cylinders facing each other for s/D > 3 is determined to be
) 2 / ( 2 2 2 1 surface on String 2 strings Uncrossed strings Crossed 2 2 2 1 D s D s F π − + = × − =
∑
∑
− or D s D s F π ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − = − 2 2 2 1 213-18 View factors from the very long grooves shown in the figure to the surroundings are to be determined. Assumptions 1 The surfaces are diffuse emitters and reflectors. 2 End effects are neglected.
Analysis (a) We designate the circular dome surface by (1) and the imaginary flat top surface by (2). Noting that (2) is flat,
0 22 = F (1) D (2) 1 1 : rule summation F21+F22 = ⎯⎯→F21= 0.64 = = = = ⎯→ ⎯ = π π (1) 2 2 A : rule y reciprocit 21 1 2 12 21 2 12 1 A F DD A F F A F
(b) We designate the two identical surfaces of length b by (1) and (3), and the imaginary flat top surface by (2). Noting that (2) is flat, 0 22 = F (3) (1) (2) b b a 5 . 0 1 : rule summation F21+F22 +F23 = ⎯⎯→F21 =F23 = (symmetry) 1 1 : rule summation F22+F2→(1+3) = ⎯⎯→F2→(1+3)= 2b a = = = ⎯→ ⎯ = + → + → + → + + + → ) 1 ( A : rule y reciprocit ) 3 1 ( 2 ) 3 1 ( 2 ) 3 1 ( 2 ) 3 1 ( ) 3 1 ( ) 3 1 ( 2 2 A A F F F A F surr (1) a b b (2) (3) (4)
(c) We designate the bottom surface by (1), the side surfaces by (2) and (3), and the imaginary top surface by (4). Surface 4 is flat and is completely surrounded by other surfaces. Therefore, F44 =0 and F4→(1+2+3) =1.
2b + a a = = = ⎯→ ⎯ = + + → + + → + + → + + + + + + → ) 1 ( A : rule y reciprocit ) 3 2 1 ( 4 ) 3 2 1 ( 4 ) 3 2 1 ( 4 ) 3 2 1 ( ) 3 2 1 ( ) 3 2 1 ( 4 4 A A F F F A F surr
13-19 A circular cone is positioned on a common axis with a circular disk, the values of F11 and F12 for specified L and D are
to be determined.
Assumptions 1 The surfaces are diffuse emitters and reflectors. 2 The surface A1 is treated as a single surface.
Analysis The area for A1, A2, and A3 are
4 5 2 2 2 2 / 1 2 2 1 πD L D D π A = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = and 4 2 3 2 A D A = =π
The surfaces A1 and A3 can be treated as an enclosure, and using the summation rule yields
1 13 11+ F =
F
Applying reciprocity relation between A1 and A3, we have
→ 31 3 13 1F A F A = F11 =1−F13 =1−(A3/A1)F31 Note that → (since 1 33 31+ F = F F31 =1 F33 =0)
Hence, the view factor F11 is
0.553 = − = − = π π 2 2 1 3 11 5 1 ) / ( 1 D D A A F
The radiation leaving A2 is intercepted by A1 and A3 equally,
23 21 F
F =
Applying reciprocity relation between A1 and A2, we have
→ 21 2 12 1F A F A = 5 ) / ( ) / ( 2 1 21 2 1 23 23 12 F F A A F A A F = = =
For coaxial parallel disks, the view factor F23 is evaluated from Table 13-1 as
1716 . 0 ] ) 4 6 ( 6 [ 2 1 ] ) 4 ( [ 2 1 4 2 1 2 1/2 2 1/2 2 / 1 2 2 3 2 23 = − − = − − = ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − = S S D D S S F where 2 1 2 2 1= = = L = D R R R 6 4 2 1 2 1 1 2 2 1 2 2 = + = + = + + = R R R S
Hence, the view factor F12 is
0.0767 = = = 5 1716 . 0 5 23 12 F F
Discussion As long as L = D, the view factors are constant at F11 = 0.553 and F12 = 0.0767 (i.e., F11 and F12 are independent
13-20 A cylindrical surface and a disk are oriented coaxially a distance L apart. The view factor F12 between them is to be
determined.
Assumptions The surfaces are diffuse emitters and reflectors. Analysis The end surfaces A3 and A4 are treated as hypothetical
surfaces. The summation rule for surfaces facing the disk can be expressed as → 24 21 23 F F F = + F21=F23 −F24 (1) From reciprocity relation, we have
(2) 12 1 21 2F AF A =
Substituting Eq. (2) in to Eq. (1) and rearranging give → 24 23 12 2 1/ ) (A A F =F −F F12 =(A2/A1)(F23−F24) (3)
The view factors F23 and F24 can be determined from the relation in Table 13-1 by treating the two surfaces as coaxial
parallel disks of identical diameters:
2 2 2 1 4 1 1 R R F Fi→j = j→i = + − + where L r R= For surface 3 (L = 2D): 25 . 0 2 2 / 2 / = = = = D D L D L r R and 05573 . 0 25 . 0 2 1 25 . 0 4 1 1 22 23 = × + × − + = F For surface 4 (L = 4D): 0.125 4 2 / 2 / = = = = D D L D L r R and 01515 . 0 125 . 0 2 1 125 . 0 4 1 1 22 24 = × + × − + = F
Substituting these values of F23 and F24 into Eq. (3), and noting that L = 2D, the view factor between the cylindrical surface
and the disk is determined to be
0.00507 = − = − = − = − = − = ) 01515 . 0 05573 . 0 ( 8 1 ) ( 8 1 ) ( ) 2 ( 4 / ) ( 4 / ) ( 24 23 24 23 2 24 23 2 24 23 1 2 12 F F F F D D D F F DL D F F A A F π π π π
13-21 A cylindrical surface and a disk are oriented coaxially a distance L apart. The view factor F12 between them is to be
determined.
Assumptions The surfaces are diffuse emitters and reflectors. Analysis The end surfaces A3 and A4 are treated as hypothetical
surfaces. The summation rule for surfaces facing the disk can be expressed as → 24 21 23 F F F = + F21=F23 −F24 (1) From reciprocity relation, we have
(2) 12 1 21 2F AF A =
Substituting Eq. (2) in to Eq. (1) and rearranging give → 24 23 12 2 1/ ) (A A F =F −F F12 =(A2/A1)(F23−F24) (3)
The view factors F23 and F24 can be determined from the relation in Table 13-1 by treating the two surfaces as coaxial
parallel disks of identical diameters:
2 2 2 1 4 1 1 R R F Fi→j = j→i = + − + where L r R= For surface 3 (L = D): 5 . 0 2 / 2 / = = = = D D L D L r R and 1716 . 0 5 . 0 2 1 5 . 0 4 1 1 2 2 23 = × + × − + = F For surface 4 (L = 2D): 25 . 0 2 2 / 2 / = = = = D D L D L r R and 05573 . 0 25 . 0 2 1 25 . 0 4 1 1 2 2 24 = × + × − + = F
Substituting these values of F23 and F24 into Eq. (3), and noting that L = D, the view factor between the cylindrical surface
and the disk is determined to be
0.0290 = − = − = − = (0.1716 0.05573) 4 1 ) ( 4 / ) ( 23 24 2 23 24 1 2 12 A F F DDL F F A F π π
Radiation Heat Transfer between Surfaces
13-22C The two methods used in radiation analysis are the matrix and network methods. In matrix method, equations 13-34 and 13-35 give N linear algebraic equations for the determination of the N unknown radiosities for an N -surface enclosure. Once the radiosities are available, the unknown surface temperatures and heat transfer rates can be determined from these equations respectively. This method involves the use of matrices especially when there are a large number of surfaces. Therefore this method requires some knowledge of linear algebra.
The network method involves drawing a surface resistance associated with each surface of an enclosure and connecting them with space resistances. Then the radiation problem is solved by treating it as an electrical network problem where the radiation heat transfer replaces the current and the radiosity replaces the potential. The network method is not practical for enclosures with more than three or four surfaces due to the increased complexity of the network.
13-23C Radiosity is the total radiation energy leaving a surface per unit time and per unit area. Radiosity includes the emitted radiation energy as well as reflected energy. Radiosity and emitted energy are equal for blackbodies since a blackbody does not reflect any radiation.
13-24C Radiation surface resistance is given as
i i i i A R ε ε −
=1 and it represents the resistance of a surface to the emission of radiation. It is zero for black surfaces. The space resistance is the radiation resistance between two surfaces and is expressed as
ij i ij AF
R = 1
13-25C The analysis of radiation exchange between black surfaces is relatively easy because of the absence of reflection. The rate of radiation heat transfer between two surfaces in this case is expressed as Q&=A1F12σ(T14 −T24) where A1 is the surface area, F12 is the view factor, and T1 and T2 are the temperatures of two surfaces.
13-26C Some surfaces encountered in numerous practical heat transfer applications are modeled as being adiabatic as the back sides of these surfaces are well insulated and net heat transfer through these surfaces is zero. When the convection effects on the front (heat transfer) side of such a surface is negligible and steady-state conditions are reached, the surface must lose as much radiation energy as it receives. Such a surface is called reradiating surface. In radiation analysis, the surface resistance of a reradiating surface is taken to be zero since there is no heat transfer through it.
13-27 Two aligned parallel rectangles are apart by a distance of 2 m. The percentage of change in radiation heat transfer rate when the rectangles are moved apart from 2 m to 8 m is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surfaces are black. 3 Convection heat transfer is not considered. Analysis For D = 2 m, L1 = 6 m and L2 = 8 m, we have
3 2 6 1 = = D L and 4 2 8 2 = = D L
From Fig. 13-5, we get
(for D = 2m) 58 . 0 12 ≈ F
For D = 8 m, L1 = 6 m and L2 = 8 m, we have
75 . 0 8 6 1 = = D L and 1 8 8 2 = = D L
From Fig. 13-5, we get
(for D = 8 m) 165 . 0 12 ≈ F
The rate of radiation heat transfer between the two rectangles is )
( 14 24
12
12 AF T T
Q& = σ −
Hence, the percentage of change in radiation heat transfer rate is
71.6%) (or 0.716 = − = = = − = = = = − = = 58 . 0 165 . 0 58 . 0 ) m 2 ( ) m 8 ( ) m 2 ( ) m 2 ( ) m 8 ( ) m 2 ( Change % 12 12 12 12 12 12 D F D F D F D Q D Q D Q & & &
Discussion By moving the distance between the two parallel rectangles from 2 m to 8 m, there is about 72% reduction in radiation heat transfer rate.
13-28 A solid sphere is placed in an evacuated equilateral triangular enclosure. The view factor from the enclosure to the sphere and the emissivity of the enclosure are to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Convection heat transfer is not considered.
Properties The emissivity of sphere is given to be ε1 = 0.45.
T2 = 420 K ε2= ? T1 = 600 K ε1= 0.45 1 m 2 m 2 m
Analysis (a) We take the sphere to be surface 1 and the surrounding enclosure to be surface 2. The view factor from surface 2 to surface 1 is determined from reciprocity relation:
0.605 = = = = − = − = = = = 21 21 21 2 12 1 2 2 2 2 2 2 2 2 2 1 ) 196 . 5 ( ) 1 )( 142 . 3 ( m 196 . 5 2 m 2 m) 1 ( m) 2 ( 3 2 3 m 142 . 3 m) 1 ( F F F A F A L D L A D A π π 2 m
(b) The net rate of radiation heat transfer can be expressed for this two-surface enclosure to yield the emissivity of the enclosure:
(
)
(
) (
)
[
]
0.150 = − + + − − ⋅ × = − + + − − = − 2 2 2 2 2 2 4 4 4 2 8 2 2 2 12 1 1 1 1 4 2 4 1 ) m 196 . 5 ( 1 ) 1 )( m 142 . 3 ( 1 ) 45 . 0 )( m 142 . 3 ( 45 . 0 1 K 420 K 600 ) K W/m 10 67 . 5 ( W 3100 1 1 1 ε ε ε ε ε ε ε σ A F A A T T Q&13-29 A long cylindrical rod coated with a new material is placed in an evacuated long cylindrical enclosure which is maintained at a uniform temperature. The emissivity of the coating on the rod is to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray.
Vacuum D1 = 0.01 m T1 = 600 K ε1= ? D2 = 0.1 m T2 = 200 K ε2 = 0.95
Properties The emissivity of the enclosure is given to be ε2 = 0.95.
Analysis The emissivity of the coating on the rod is determined from
(
) (
)
⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − + − ⋅ × = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − + − = − 10 1 95 . 0 95 . 0 1 1 ] K 200 K 600 )[ K W/m 10 67 . 5 ]( m) m)(1 01 . 0 ( [ W 12 1 1 ) ( 1 4 4 4 2 8 2 1 2 2 1 4 2 4 1 1 12 ε π ε ε ε σ r r T T A Q& which gives ε1 = 0.052713-30E Top and side surfaces of a cubical furnace are black, and are maintained at uniform temperatures. Net radiation heat transfer rate to the base from the top and side surfaces are to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Convection heat transfer is not considered.
Properties The emissivities are given to be ε = 0.4 for the bottom surface and 1 for other surfaces.
Analysis We consider the base surface to be surface 1, the top surface to be surface 2 and the side surfaces to be surface 3. The cubical furnace can be considered to be three-surface enclosure. The areas and blackbody emissive powers of surfaces are 2 2 3 2 2 2 1=A =(10ft) =100ft A =4(10ft) =400ft A T3 = 2400 R ε3 = 1 T2 = 1600 R ε2 = 1 2 4 4 2 8 4 3 3 2 4 4 2 8 4 2 2 2 4 4 2 8 4 1 1 Btu/h.ft 866 , 56 ) R 2400 )( R . Btu/h.ft 10 1714 . 0 ( Btu/h.ft 233 , 11 ) R 1600 )( R . Btu/h.ft 10 1714 . 0 ( Btu/h.ft 702 ) R 800 )( R . Btu/h.ft 10 1714 . 0 ( = × = = = × = = = × = = − − − T E T E T E b b b σ σ σ The view factor from the base to the top surface of the cube is F12 =0.2. From the summation rule, the view factor from the base or top to the side surfaces is
T1 = 800 R ε1 = 0.4 8 . 0 2 . 0 1 1 1 13 12 13 12 11+F +F = ⎯⎯→F = −F = − = F
since the base surface is flat and thus F11=0. Then the radiation resistances become
2 -2 13 1 13 2 -2 12 1 12 2 -2 1 1 1 1 ft 0125 . 0 ) 8 . 0 )( ft 100 ( 1 1 ft 0500 . 0 ) 2 . 0 )( ft 100 ( 1 1 ft 015 . 0 ) 4 . 0 )( ft 100 ( 4 . 0 1 1 = = = = = = = − = − = F A R F A R A R ε ε
Note that the side and the top surfaces are black, and thus their radiosities are equal to their emissive powers. The radiosity of the base surface is determined
0 13 1 3 12 1 2 1 1 1− + − + − = R J E R J E R J Eb b b Substituting, 2 1 1 1 1 0 28,925 W/m 0125 . 0 866 , 56 05 . 0 233 , 11 015 . 0 702 − = ⎯⎯→ = + − + − J J J J
(a) The net rate of radiation heat transfer between the base and the side surfaces is Btu/h 10 2.235× 6 = − = − = -2 2 13 1 3 31 ft 0125 . 0 Btu/h.ft ) 915 , 28 866 , 56 ( R J E Q& b
(b) The net rate of radiation heat transfer between the base and the top surfaces is Btu/h 10 3.538× 5 = − = − = -2 2 12 2 1 12 ft 05 . 0 Btu/h.ft ) 233 , 11 925 , 28 ( R E J Q& b
The net rate of radiation heat transfer to the base surface is finally determined from Btu/h 10 1.882× 6 = × + × − = + = 5 6 31 21 1 Q Q 3.538 10 2.235 10
Q& & &
Discussion The same result can be found form
Btu/h 10 882 . 1 ft 015 . 0 Btu/h.ft ) 702 925 , 28 ( 6 2 -2 1 1 1 1 = − = × − = R E J Q& b
13-31E Prob. 13-30E is reconsidered. The effect of base surface emissivity on the net rates of radiation heat transfer between the base and the side surfaces, between the base and top surfaces, and to the base surface is to be investigated. Analysis The problem is solved using EES, and the solution is given below.
"GIVEN" a=10 [ft] epsilon_1=0.4 T_1=800 [R] T_2=1600 [R] T_3=2400 [R] "ANALYSIS"
sigma=0.1714E-8 [Btu/h-ft^2-R^4] “Stefan-Boltzmann constant"
"Consider the base surface 1, the top surface 2, and the side surface 3"
E_b1=sigma*T_1^4 E_b2=sigma*T_2^4 E_b3=sigma*T_3^4 A_1=a^2 A_2=A_1 A_3=4*a^2
F_12=0.2 "view factor from the base to the top of a cube"
F_11+F_12+F_13=1 "summation rule"
F_11=0 "since the base surface is flat"
R_1=(1-epsilon_1)/(A_1*epsilon_1) "surface resistance"
R_12=1/(A_1*F_12) "space resistance"
R_13=1/(A_1*F_13) "space resistance"
(E_b1-J_1)/R_1+(E_b2-J_1)/R_12+(E_b3-J_1)/R_13=0 "J_1 : radiosity of base surface" "(a)" Q_dot_31=(E_b3-J_1)/R_13 "(b)" Q_dot_12=(J_1-E_b2)/R_12 Q_dot_21=-Q_dot_12 Q_dot_1=Q_dot_21+Q_dot_31 ε1 Q& 31 [Btu/h] 12 Q& [Btu/h] 1 Q& [Btu/h] 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 1.106E+06 1.295E+06 1.483E+06 1.671E+06 1.859E+06 2.047E+06 2.235E+06 2.423E+06 2.612E+06 2.800E+06 2.988E+06 3.176E+06 3.364E+06 3.552E+06 3.741E+06 3.929E+06 4.117E+06 636061 589024 541986 494948 447911 400873 353835 306798 259760 212722 165685 118647 71610 24572 -22466 -69503 -116541 470376 705565 940753 1.176E+06 1.411E+06 1.646E+06 1.882E+06 2.117E+06 2.352E+06 2.587E+06 2.822E+06 3.057E+06 3.293E+06 3.528E+06 3.763E+06 3.998E+06 4.233E+06
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0x106 1.5x106 2.0x106 2.5x106 3.0x106 3.5x106 4.0x106 4.5x106 ε1 Q31 [B tu /h ] 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 -200000 -100000 0 100000 200000 300000 400000 500000 600000 700000 ε1 Q12 [Bt u /h ] 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0.0x100 5.0x105 1.0x106 1.5x106 2.0x106 2.5x106 3.0x106 3.5x106 4.0x106 4.5x106 ε1 Q1 [B tu /h ]
13-32 Two very large parallel plates are maintained at uniform temperatures. The net rate of radiation heat transfer between the two plates is to be determined.
T1 = 600 K
ε1 = 0.5
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Convection heat transfer is not considered.
Properties The emissivities ε of the plates are given to be 0.5 and 0.9.
T2 = 400 K
ε2 = 0.9
Analysis The net rate of radiation heat transfer between the two surfaces per unit area of the plates is determined directly from 2 W/m 2793 = − + − ⋅ × = − + − = − 1 9 . 0 1 5 . 0 1 ] ) K 400 ( ) K 600 )[( K W/m 10 67 . 5 ( 1 1 1 ) ( 8 2 4 4 4 2 1 4 2 4 1 12 ε ε σ T T A Q s &
13-33 Prob. 13-32 is reconsidered. The effects of the temperature and the emissivity of the hot plate on the net rate of radiation heat transfer between the plates are to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
"GIVEN"
T_1=600 [K] T_2=400 [K] epsilon_1=0.5 epsilon_2=0.9
sigma=5.67E-8 [W/m^2-K^4] “Stefan-Boltzmann constant" "ANALYSIS" q_dot_12=(sigma*(T_1^4-T_2^4))/(1/epsilon_1+1/epsilon_2-1) T1 [K] q&12 [W/m2] 500 991.1 525 1353 550 1770 575 2248 600 2793 625 3411 650 4107 675 4888 700 5761 725 6733 750 7810 775 9001 800 10313 825 11754 850 13332 875 15056 900 16934 925 18975 950 21188 975 23584 1000 26170 500 600 700 800 900 1000 0 5000 10000 15000 20000 25000 30000 T1 [K] q12 [ W /m 2 ] ε1 q&12 [W/m2] 0.1 583.2 0.15 870 0.2 1154 0.25 1434 0.3 1712 0.35 1987 0.4 2258 0.45 2527 0.5 2793 0.55 3056 0.6 3317 0.65 3575 0.7 3830 0.75 4082 0.8 4332 0.85 4580 0.9 4825 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 ε1 q12 [ W /m 2 ]
13-34 The radiation heat flux between two infinitely long parallel plates of specified surface temperatures is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surfaces are black. 3 Convection heat transfer is not considered. 4 The surface temperatures are uniform.
Analysis From the Hottel’s crossed-strings method, we have
) surface on String ( 2 ) strings Uncrossed ( ) strings Crossed ( i Fi j × Σ − Σ = →
For uncrossed strings, we have
w w w w w L L ( 2 2)1/2 ( 2 2)1/2 2 2 1= = + = + =
For crossed strings, we have
w w
w
L3 =( 2+4 2)1/2= 5 and L4 =w
Applying the Hottel’s crossed-strings method, we get F12 as
204 . 0 2 ) 2 2 ( ) 5 ( 2 ) ( ) ( 3 4 1 2 12 = + − + = + − + = w w w w w w L L L L F
The radiation heat flux between the two surfaces is
2 W/m 2680 = − ⋅ × = − = −8 2 4 4 4 4 4 2 4 1 12 12 K ) 300 700 )( K W/m 10 67 . 5 )( 204 . 0 ( ) (T T F q& σ
Discussion The Hottel’s crossed-string method is applicable only to surfaces that are very long, such that they can be considered to be two-dimensional and radiation interaction through the end surfaces is negligible.
13-35 The base and the dome of a long semi-cylindrical dryer are maintained at uniform temperatures. The drying rate per unit length experienced by the base surface is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surfaces are black. 3 Convection heat transfer is not considered. 4 The dryer is well insulated from heat loss to the surrounding.
Properties The latent heat of vaporization for water is hfg = 2257 kJ/kg (Table A-2)
Analysis The view factor from the dome to the base is determined from → (where
1 12 11+ F =
F F12 =1 F11=0)
Hence, from reciprocity relation, we get π π 2 2 / 12 2 1 21= A F = DLDL = A F
Applying energy balance on the base surface, we have
fg
h m Q Q&21 = &evap= &
Hence, ) ( 2 ) ( 24 14 21 24 14 21 2F T T DLF T T A h m& fg = σ − =π σ − or m kg/s 0.0370 ⋅ = − ⋅ × × = − = − = − = −8 2 4 4 4 4 3 4 1 4 2 4 1 4 2 4 1 4 2 21 K ) 370 1000 )( K W/m 10 67 . 5 ( ) J/kg 10 2257 ( m) 5 . 1 ( ) ( ) ( 2 2 ) ( 2 T T h D T T h D T T F h D L m fg fg fg σ σ π π σ π &
Discussion The view factor from the dome to the base is constant F21 = 2/π, which implies that the view factor is
13-36 The base, top, and side surfaces of a furnace of cylindrical shape are black, and are maintained at uniform temperatures. The net rate of radiation heat transfer to or from the top surface is to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are black. 3 Convection heat transfer is not considered.
H = 3 m T 3 = 500 K ε3 = 1 T2 = 1400 K ε2 = 1 R = 3 m T1 = 700 K ε1 = 1 R = 3 m
Properties The emissivity of all surfaces are ε = 1 since they are black. Analysis We consider the top surface to be surface 1, the base surface to be surface 2 and the side surfaces to be surface 3. The cylindrical furnace can be considered to be three-surface enclosure. We assume that steady-state conditions exist. Since all surfaces are black, the radiosities are equal to the emissive power of surfaces, and the net rate of radiation heat transfer from the top surface can be determined from
) ( ) ( 14 24 1 13 14 34 12 1F T T AF T T A Q&= σ − + σ − and A1=πR2=π(3m)2 =28.27m2
The view factor from the base to the top surface of the cylinder is F12=0.38 (From Figure 13-7). The view factor from the
base to the side surfaces is determined by applying the summation rule to be
62 . 0 38 . 0 1 1 1 13 12 13 12 11+F +F = ⎯⎯→F = −F = − = F Substituting, kW 3579 − = × − = × + × = − + − = W 10 579 . 3 ) K 1400 -K )(700 .K W/m 10 67 . 5 )( 62 . 0 )( m 27 . 28 ( ) K 500 -K )(700 .K W/m 10 67 )(0.38)(5. m (28.27 ) ( ) ( 6 4 4 4 2 8 -2 4 4 4 2 8 -2 4 3 4 1 13 1 4 2 4 1 12 1F T T AF T T A Q& σ σ
13-37E A room is heated by electric resistance heaters placed on the ceiling which is maintained at a uniform temperature. The rate of heat loss from the room through the floor is to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Convection heat transfer is not considered. 4 There is no heat loss through the side surfaces.
Properties The emissivities are ε = 1 for the ceiling and ε = 0.8 for the floor. The emissivity of insulated (or reradiating) surfaces is also 1.
Analysis The room can be considered to be three-surface enclosure with the ceiling surface 1, the floor surface 2 and the side surfaces surface 3. We assume steady-state conditions exist. Since the side surfaces are reradiating, there is no heat transfer through them, and the entire heat lost by the ceiling must be gained by the floor. Then the rate of heat loss from the room through its floor can be determined from
Insulated side surfacess Ceiling: 12 ft × 12 ft T1 = 90°F ε1 = 1 2 1 23 13 12 2 1 1 1 1 R R R R E E Q b b + ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + + − = − & 9 ft where T2 = 65°F ε2 = 0.8 2 4 4 2 8 4 2 2 2 4 4 2 8 4 1 1 Btu/h.ft 130 ) R 460 65 )( R . Btu/h.ft 10 1714 . 0 ( Btu/h.ft 157 ) R 460 90 )( R . Btu/h.ft 10 1714 . 0 ( = + × = = = + × = = − − T E T E b b σ σ and 2 2 2 1= A =(12ft) =144ft A
The view factor from the floor to the ceiling of the room is F12 =0.27 (From Figure 13-5). The view factor from the ceiling
or the floor to the side surfaces is determined by applying the summation rule to be
73 . 0 27 . 0 1 1 1 13 12 13 12 11+F +F = ⎯⎯→F = −F = − = F
since the ceiling is flat and thus F11 =0. Then the radiation resistances which appear in the equation above become
2 -2 13 1 23 13 2 -2 12 1 12 2 -2 2 2 2 2 ft 009513 . 0 ) 73 . 0 )( ft 144 ( 1 1 ft 02572 . 0 ) 27 . 0 )( ft 144 ( 1 1 ft 00174 . 0 ) 8 . 0 )( ft 144 ( 8 . 0 1 1 = = = = = = = = − = − = F A R R F A R A R ε ε Substituting, Btu/h 2130 = + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − = − 2 -1 2 -2 -2 12 ft 00174 . 0 ) ft 009513 . 0 ( 2 1 ft 02572 . 0 1 Btu/h.ft ) 130 157 ( Q&
13-38 Two very long concentric cylinders are maintained at uniform temperatures. The net rate of radiation heat transfer between the two cylinders is to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Convection heat transfer is not considered.
Vacuum D1 = 0.35 m T1 = 950 K ε1= 1 D2 = 0.5 m T2 = 500 K ε2 = 0.55
Properties The emissivities of surfaces are given to be ε1 = 1 and ε2 = 0.55.
Analysis The net rate of radiation heat transfer between the two cylinders per unit length of the cylinders is determined from
kW 29.81 = = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − + − ⋅ × = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − + − = − W 810 , 29 5 5 . 3 55 . 0 55 . 0 1 1 1 ] ) K 500 ( K) 950 )[( K W/m 10 67 . 5 ]( m) m)(1 35 . 0 ( [ 1 1 ) ( 4 4 4 2 8 2 1 2 2 1 4 2 4 1 1 12 π ε ε ε σ r r T T A Q&
13-39 Radiation heat transfer occurs between a sphere and a circular disk. The view factors and the net rate of radiation heat transfer for the existing and modified cases are to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Convection heat transfer is not considered.
Properties The emissivities of sphere and disk are given to be ε1 = 0.9 and ε2 = 0.5, respectively.
Analysis (a) We take the sphere to be surface 1 and the disk to be surface 2. The view factor from surface 1 to surface 2 is determined from 0.2764 = ⎪⎭ ⎪ ⎬ ⎫ ⎪⎩ ⎪ ⎨ ⎧ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − = ⎪⎭ ⎪ ⎬ ⎫ ⎪⎩ ⎪ ⎨ ⎧ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + − = − −0.5 2 0.5 2 2 12 0.60m m 2 . 1 1 1 5 . 0 1 1 5 . 0 h r F
The view factor from surface 2 to surface 1 is determined from reciprocity relation: h r2 r1 0.0691 = = = = = = = = = 21 21 21 2 12 1 2 2 2 2 2 2 2 2 1 1 ) 524 . 4 ( ) 2764 . 0 )( 131 . 1 ( m 524 . 4 m) 2 . 1 ( m 131 . 1 m) 3 . 0 ( 4 4 F F F A F A r A r A π π π π
(b) The net rate of radiation heat transfer between the surfaces can be determined from
(
)
[
(
) (
)
]
=8550 W − + + − − ⋅ × = − + + − − = − ) 5 . 0 )( m 524 . 4 ( 5 . 0 1 ) 2764 . 0 )( m 131 . 1 ( 1 ) 9 . 0 )( m 131 . 1 ( 9 . 0 1 K 473 K 873 ) K W/m 10 67 . 5 ( 1 1 1 2 2 2 4 4 4 2 8 2 2 2 12 1 1 1 1 4 2 4 1 ε ε ε ε σ A F A A T T Q&(c) The best values are ε1=ε2 =1 and h=r1=0.3m. Then the view factor becomes
3787 . 0 m 30 . 0 m 2 . 1 1 1 5 . 0 1 1 5 . 0 5 . 0 2 5 . 0 2 2 12 = ⎪⎭ ⎪ ⎬ ⎫ ⎪⎩ ⎪ ⎨ ⎧ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − = ⎪⎭ ⎪ ⎬ ⎫ ⎪⎩ ⎪ ⎨ ⎧ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + − = − − h r F
The net rate of radiation heat transfer in this case is
13-40 Two parallel disks whose back sides are insulated are black, and are maintained at a uniform temperature. The net rate of radiation heat transfer from the disks to the environment is to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Convection heat transfer is not considered.
Properties The emissivities of all surfaces are ε = 1 since they are black. Analysis Both disks possess same properties and they are black. Noting that environment can also be considered to be blackbody, we can treat this geometry as a three surface enclosure. We consider the two disks to be surfaces 1 and 2 and the environment to be surface 3. Then from Figure 13-7, we read ) rule summation ( 74 . 0 26 . 0 1 26 . 0 13 21 12 = − = = = F F F
The net rate of radiation heat transfer from the disks into the environment then becomes
(
) (
)
W 781 = − ⋅ × = − = = + = − W/m K )[450K 300K ] 10 67 . 5 ]( ) m 3 . 0 ( )[ 74 . 0 ( 2 ) ( 2 2 4 4 4 2 8 2 4 3 4 1 1 13 3 13 23 13 3 π σ T T A F Q Q Q Q Q & & & & & 0.40 m Disk 2, T2 = 450 K, ε2 = 1 Disk 1, T1 = 450 K, ε1 = 1 D = 0.6 m Environment T3 =300 K ε1 = 113-41E Two black parallel rectangles are spaced apart by a distance of 1 ft, the temperature of the bottom rectangle is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surfaces are black. 3 Convection heat transfer is not considered. Analysis For D = 1 ft, L1 = 3 ft and L2 = 5 ft, we have
3 1 3 1 = = D L and 5 1 5 2 = = D L
From Fig. 13-5, we get
60 . 0 12 ≈
F
The rate of radiation heat transfer between the two rectangles is ) ( ) ( 14 24 1 2 12 14 24 12 12 AF T T L L F T T Q& = σ − = σ − Hence 4 / 1 4 4 4 2 8 4 / 1 4 2 12 2 1 12 1 R ) 460 60 ( ) R ft Btu/h 10 1714 . 0 )( 60 . 0 )( ft 5 )( ft 3 ( Btu/h 180000 ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ + + ⋅ ⋅ × = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + = − T F L L Q T σ & R 1851 = 1 T
Discussion If T1 and T2 are constant, increasing the distance between the two rectangles will decrease the view factor F12,
13-42 Two coaxial parallel disks of equal diameter 1 m are originally placed at a distance of 1 m apart. The new distance between the disks such that there is a 75% reduction in radiation heat transfer rate from the original distance of 1 m is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surfaces are black. 3 Convection heat transfer is not considered. Analysis For coaxial parallel disks, from Table 13-1, with i = 1, j = 2,
] ) 4 ( [ 2 1 4 2 1 2 1/2 2 / 1 2 1 2 2 12 = − − ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − = S S D D S S F (1) 2 2 2 1 2 2 2 1 2 4( / ) 1 1 L D R R R S = + + = + = + (2) where L D R R R 2 2 1= = =
Substituting Eq. (2) into Eq. (1), we get
} ] 1 ) / )[( / ( 4 ) / ( 4 2 { 2 1 } ] ) / ( 16 ) / ( 16 [ ) / ( 4 2 { 2 1 ) } 4 ] ) / ( 4 2 {[ ) / ( 4 2 ( 2 1 2 / 1 2 2 2 / 1 2 4 2 2 / 1 2 2 2 12 + − + = + − + = − + − + = D L D L D L D L D L D L D L D L F 2 / 1 2 2 12=1+2(L/D) −2(L/D)[(L/D) +1] F
Hence, for D = 1 m we have
(3) 2 / 1 2 2 12=1+2L −2L[L +1] F
From Eq. (3), the view factor at the original distance L = 1 m (with D = l m) is
1716 . 0 2 2 3 ] 1 ) 1 )[( 1 ( 2 ) 1 ( 2 1 2 2 1/2 old , 12 = + − + = − = F
The rate of radiation heat transfer between the two surfaces is )
( 14 24
12
12 AF T T
Q& = σ −
The percentage of reduction in radiation heat transfer rate can be expressed as
old , 12 new , 12 old , 12 old , 12 new , 12 old , 12 Change % F F F Q Q Q − = − = & & &
For the two surfaces to experience 75% reduction in radiation heat transfer rate, the new view factor should be
0429 . 0 25 . 0 12 ,old new , 12 = F = F
Then, substituting the value of
F
12 ,new into Eq. (3), the new distance Lnew such that the two surfaces experience 75%reduction in radiation heat transfer rate can be calculated as
2 / 1 2 new new 2 new 2 / 1 2 new new 2 new new , 12 ] 1 [ 2 2 1 0429 . 0 ] 1 [ 2 2 1 + − + = + − + = L L L L L L F Hence, Lnew =2.31m
Discussion Increasing the distance between the disks would decrease the view factor F12, thereby reducing the radiation heat
13-43 A furnace shaped like a long equilateral-triangular duct is considered. The temperature of the base surface is to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Convection heat transfer is not considered. 4 End effects are neglected.
Properties The emissivities of surfaces are given to be ε1 = 0.8 and ε2 = 0.4.
q1 = 800 W/m2
ε1 = 0.8
T2 = 600 K
ε2 = 0.4
b = 2 m
Analysis This geometry can be treated as a two surface enclosure since two surfaces have identical properties. We consider base surface to be surface 1 and other two surface to be surface 2. Then the view factor between the two becomes . The temperature of the base surface is determined from 1 12 = F
( ) (
)
K 630 = − + + − − ⋅ × = − + + − − = − 1 2 2 2 4 4 1 4 2 8 2 2 2 12 1 1 1 1 4 2 4 1 12 ) 4 . 0 ( ) m 2 ( 4 . 0 1 ) 1 )( m 1 ( 1 ) 8 . 0 )( m 1 ( 8 . 0 1 ] K 600 )[ K W/m 10 67 . 5 ( W 800 1 1 1 ) ( T T A F A A T T Q ε ε ε ε σ &13-44 Prob. 13-43 is reconsidered. The effects of the rate of the heat transfer at the base surface and the temperature of the side surfaces on the temperature of the base surface are to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
"GIVEN" a=2 [m] epsilon_1=0.8 epsilon_2=0.4 Q_dot_12=800 [W] T_2=600 [K] sigma=5.67E-8 [W/m^2-K^4] "ANALYSIS"
"Consider the base surface to be surface 1, the side surfaces to be surface 2"
Q_dot_12=(sigma*(T_1^4-T_2^4))/((1-epsilon_1)/(A_1*epsilon_1)+1/(A_1*F_12)+(1-epsilon_2)/(A_2*epsilon_2))
F_12=1
A_1=1 "[m^2], since rate of heat supply is given per meter square area"
A_2=2*A_1 12 Q& [W] T1 [K] 500 525 550 575 600 625 650 675 700 725 750 775 800 825 850 875 900 925 950 975 1000 619.4 620.4 621.3 622.2 623.1 624 624.9 625.8 626.7 627.6 628.5 629.4 630.3 631.2 632 632.9 633.8 634.6 635.5 636.4 637.2 500 600 700 800 900 1000 617.5 621.5 625.5 629.5 633.5 637.5 Q12 [W] T1 [ K ]
T2 [K] T1 [K] 300 325 350 375 400 425 450 475 500 525 550 575 600 625 650 675 700 436.5 445.5 456 468.1 481.7 496.7 512.9 530.4 548.8 568.1 588.2 609 630.3 652.1 674.3 696.9 719.7 300 350 400 450 500 550 600 650 700 400 450 500 550 600 650 700 750 T2 [K] T1 [ K ]
13-45 The floor and the ceiling of a cubical furnace are maintained at uniform temperatures. The net rate of radiation heat transfer between the floor and the ceiling is to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Convection heat transfer is not considered.
Properties The emissivities of all surfaces are ε = 1 since they are black or reradiating.
Analysis We consider the ceiling to be surface 1, the floor to be surface 2 and the side surfaces to be surface 3. The furnace can be considered to be three-surface enclosure. We assume that steady-state conditions exist. Since the side surfaces are reradiating, there is no heat transfer through them, and the entire heat lost by the ceiling must be gained by the floor. The view factor from the ceiling to the floor of the furnace is F12=0.2. Then the rate of heat loss from the ceiling can be determined from 1 23 13 12 2 1 1 1 1 − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + + − = R R R E E Q& b b a = 4 m Reradiating side surfacess T1 = 1100 K ε1 = 1 where 2 4 4 2 8 4 2 2 2 4 4 2 8 4 1 1 W/m 5188 ) K 550 )( K . W/m 10 67 . 5 ( W/m 015 , 83 ) K 1100 )( K . W/m 10 67 . 5 ( = × = = = × = = − − T E T E b b σ σ and T2 = 550 K ε2 = 1 2 2 2 1= A =(4m) =16m A 2 -2 13 1 23 13 2 -2 12 1 12 m 078125 . 0 ) 8 . 0 )( m 16 ( 1 1 m 3125 . 0 ) 2 . 0 )( m 16 ( 1 1 = = = = = = = F A R R F A R Substituting, kW 747 = × = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − = − 7.47 10 W ) m 078125 . 0 ( 2 1 m 3125 . 0 1 W/m ) 5188 015 , 83 ( 5 1 2 -2 -2 12 Q&
13-46 Two concentric spheres are maintained at uniform temperatures. The net rate of radiation heat transfer between the two spheres and the convection heat transfer coefficient at the outer surface are to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray.
Properties The emissivities of surfaces are given to be ε1 = 0.5
and ε2 = 0.7.
Analysis The net rate of radiation heat transfer between the two spheres is
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W 2641 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − + − ⋅ × = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − + − = − 2 4 4 4 2 8 2 2 2 2 1 2 2 1 4 2 4 1 1 12 m 3 . 0 m 15 . 0 7 . 0 7 . 0 1 5 . 0 1 K 500 K 800 K W/m 10 67 . 5 m) 3 . 0 ( 1 1 π ε ε ε σ r r T T A Q& ε= 0.35 D1 = 0.3 m T1 = 800 K ε1= 0.5 D2 = 0.6 m T2 = 500 K ε2 = 0.7 Tsurr =30°C T∞ = 30°CRadiation heat transfer rate from the outer sphere to the surrounding surfaces are
W 1214 ] ) K 273 30 ( ) K 500 )[( K W/m 10 67 . 5 ]( m) 6 . 0 ( )[ 1 )( 35 . 0 ( ) ( 4 4 4 2 8 2 4 4 2 2 = + − ⋅ × = − = − π σ ε surr rad FA T T Q&
The convection heat transfer rate at the outer surface of the cylinder is determined from requirement that heat transferred from the inner sphere to the outer sphere must be equal to the heat transfer from the outer surface of the outer sphere to the environment by convection and radiation. That is,
W 1427 1214 2641 12− = − = = rad conv Q Q
Q& & &
Then the convection heat transfer coefficient becomes
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C W/m 6.40 2⋅° = = − = ∞ h h T T hA Qconv K) 303 -K (500 m) 6 . 0 ( W 1427 2 2 2 . π &13-47 A spherical tank filled with liquid nitrogen is kept in an evacuated cubic enclosure. The net rate of radiation heat transfer to the liquid nitrogen is to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Convection heat transfer is not considered. 4 The thermal resistance of the tank is negligible.
Properties The emissivities of surfaces are given to be ε1 = 0.1 and ε2 = 0.8.
Analysis We take the sphere to be surface 1 and the surrounding cubic enclosure to be surface 2. Noting that , for this two-surface enclosure, the net rate of radiation heat transfer to liquid nitrogen can be determined from
1 12 = F
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W 228 = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − + − ⋅ × − = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − + − − = − = − 2 2 4 4 4 2 8 2 2 1 2 2 1 4 2 4 1 1 12 21 m) 6(3 m) 2 ( 8 . 0 8 . 0 1 1 . 0 1 K 240 K 100 K W/m 10 67 . 5 m) 2 ( 1 1 π π ε ε ε σ A A T T A Q Q& & Liquid N2 Vacuum Cube, a =3 m T2 = 240 K ε2 = 0.8 D1 = 2 m T1 = 100 K ε1= 0.113-48 A spherical tank filled with liquid nitrogen is kept in an evacuated spherical enclosure. The net rate of radiation heat transfer to the liquid nitrogen is to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Convection heat transfer is not considered. 4 The thermal resistance of the tank is negligible.
Properties The emissivities of surfaces are given to be ε1 = 0.1 and
ε2 = 0.8.
Analysis The net rate of radiation heat transfer to liquid nitrogen can be determined from
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W 227 = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − + − ⋅ × = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − + − = − 2 2 4 4 4 2 8 2 2 2 2 1 2 2 1 4 2 4 1 1 12 m) (1.5 m) 1 ( 8 . 0 8 . 0 1 1 . 0 1 K 100 K 240 K W/m 10 67 . 5 m) 2 ( 1 1 π ε ε ε σ r r T T A Q& Liquid N2 Vacuum D1 = 2 m T1 = 100 K ε1 = 0.1 D2 = 3 m T2 = 240 K ε2 = 0.813-49 Prob. 13-47 is reconsidered. The effects of the side length and the emissivity of the cubic enclosure, and the emissivity of the spherical tank on the net rate of radiation heat transfer are to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
"GIVEN" D=2 [m] a=3 [m] T_1=100 [K] T_2=240 [K] epsilon_1=0.1 epsilon_2=0.8
sigma=5.67E-8 [W/m^2-K^4] “Stefan-Boltzmann constant" "ANALYSIS"
"Consider the sphere to be surface 1, the surrounding cubic enclosure to be surface 2"
Q_dot_12=(A_1*sigma*(T_1^4-T_2^4))/(1/epsilon_1+(1-epsilon_2)/epsilon_2*(A_1/A_2)) Q_dot_21=-Q_dot_12 A_1=pi*D^2 A_2=6*a^2 a [m] [W] Q& 21 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 226.3 227.4 227.9 228.3 228.5 228.7 228.8 228.9 228.9 229 229 229.1 229.1 229.1 229.1 229.1 229.1 229.2 229.2 229.2 229.2 2 4 6 8 10 12 226 226.5 227 227.5 228 228.5 229 229.5 a [m] Q21 [W ]
ε1 Q& 21 [W] 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 227.9 340.9 453.3 565 676 786.4 896.2 1005 1114 1222 1329 1436 1542 1648 1753 1857 1961 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 200 400 600 800 1000 1200 1400 1600 1800 2000